The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 179 ms)
↳4 CdtProblem
↳5 CdtInstantiationProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳8 CdtProblem
↳9 CdtUsableRulesProof (⇔, 0 ms)
↳10 CdtProblem
↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 8 ms)
↳12 CdtProblem
↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳14 BOUNDS(1, 1)

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), s(y)) → f(x, s(c(s(y))))

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
S tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

(3) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
We considered the (Usable) Rules:none
And the Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = x2   
POL(c(x1)) = [4] + x1   
POL(c1(x1, x2)) = x1 + x2   
POL(c2(x1)) = x1   
POL(f(x1, x2)) = [3] + [2]x1   
POL(s(x1)) = 0   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
S tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

(5) CdtInstantiationProof (BOTH BOUNDS(ID, ID) transformation)

Use instantiation to replace F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) by

F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
S tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:

F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))
Tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(z0, c(z1)) → f(z0, s(f(z1, z1)))
f(s(z0), s(z1)) → f(z0, s(c(s(z1))))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = [4]x1 + x2   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(s(x1)) = [5] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:none
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1))
F(s(z0), s(z1)) → c2(F(z0, s(c(s(z1)))))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(14) BOUNDS(1, 1)